1. ## doublte integral question

"The double integral $\iint_{R}^{} \, f(x,y) dx\,dy$ can be expressed as $\int_{\pi}^{2\pi}$ $\int_{1}^{2}$ $r^2 sin(\theta) dr\,d\Theta$ in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates?

2. Originally Posted by tennisfreak412
"The double integral $\iint_{R}^{} \, dx\,dy$ can be expressed as $\int_{Pi}^{2Pi}$ $\int_{1}^{2}$ $r^2 dr\,dTheta$
ummm...what's the question? are you supposed to find the region R?

by the way, type \pi to get $\pi$ and \theta to get $\theta$

3. sorry.. question is here:

"The double integral $\iint_{R}^{} \, f(x,y) dx\,dy$ can be expressed as $\int_{\pi}^{2\pi}$ $\int_{1}^{2}$ $r^2 sin(\theta) dr\,d\Theta$ in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates?

4. Originally Posted by tennisfreak412
"The double integral $\iint_{R}^{} \, f(x,y) dx\,dy$ can be expressed as $\int_{\pi}^{2\pi}$ $\int_{1}^{2}$ $r^2 sin(\theta) dr\,d\Theta$ in polar coordinates. What is the region R? What is the formula for f(x,y)? What is the integral in cartesian coordinates?
the radius goes from 1 to 2, the angle theta goes from pi to 2pi. thus the region R is the bottom half of the annulus formed by the circles with common center the origin, and radii 1 and 2 (see the diagram below).

since $\int \int f(x,y)~dxdy \to \int \int f(r \cos \theta, r \sin \theta)~rdrd \theta$ we see that the integral in polar coordinates is $r$. and we know that $r = \sqrt{x^2 + y^2}$

i leave the rest to you

can you continue?

5. I got that, but the difficulty I'm having is doing the double integral for the annulus. I know that I can have the dy limits going from -2 to 0, right? but then how do I do the dx limits so that they incorporate both sides of the hole in the annulus?

6. Originally Posted by tennisfreak412
I got that, but the difficulty I'm having is doing the double integral for the annulus. I know that I can have the dy limits going from -2 to 0, right? but then how do I do the dx limits so that they incorporate both sides of the hole in the annulus?
Hint:

the equation for the inner semi-circle is: $- \sqrt{1 - x^2}$

the equation for the outer semi-circle is: $- \sqrt{4 - x^2}$

can you continue?